CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    Potential energy in a spring when stretched by 2 cm is U. Its potential energy, when stretched by 10 cm is:

    A)  \[\frac{U}{25}\]                              

    B)         \[\frac{U}{5}\]

    C)  \[25U\]               

    D)         \[5U\]

    E)  none of these

    Correct Answer: C

    Solution :

    Since potential energy \[U\propto {{x}^{2}}\] Therefore, new potential energy\[={{5}^{2}}U=25U\]


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